Zadanie 10a, Paper 1, listopad 2022, Matematyka AA IB HL

$3\cos^2x=\sin^2x$ gives $\tan^2x=3$. Thus $\tan x=\pm\sqrt3$, yielding x = π/3, 2π/3 in the specified interval. Equivalently, $4\cos^2x-1=0$.

M1:reduce to one trigonometric function. A1:correct squared equation. A1:correct square-root branches. A1:π/3. A1:2π/3. Extra solutions outside the domain lose one final A1. Without working, at most the implied method mark and two final answer marks. If both otherwise correct roots are given in degrees, award at most 4/5; a single degree answer from an omitted square-root branch earns only M1A1 before the answer marks.